Question

How would I go about finding all values of x in 0 to 360 that satisfy the equation sin^2x = sinxcosx ?

Answer 1

\[\sin^2x-\sin x \cos x=0 \ \ then \ \ \sin x(\sin x-\cos x)=0 \ \ \\ so \ \ \ \sin x =0 \ \ or \ \ \sin x=\cos x \ \\ if \ \ \sin x=0 \ \ then \ \ x=0,\pi,2\pi \\ if \ \ sinx=\cos x \ \ then \ \ \tan x=1 \ \ and \ we \ have \ x=\frac{\pi}{4} \ , \ x=\frac{5\pi}{4}\]